When a specific amount of acetone (C3H6O) is added to 100.0 g of pure water at 65°C, the vapor pressure of water over the solution is lowered by 1.556 kPa. Given the vapor pressure of water at 65°C is 25.022 kPa, what is the mass of acetone added?

The mass of acetone that must be added is 4.101 grams

Explanation:

The relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The equation used to calculate relative lowering of vapor pressure follows:

where,

= relative lowering in vapor pressure  = 1.556 kPa

i = Van't Hoff factor = 1 (for non electrolytes)

= mole fraction of solute = ?

= vapor pressure of pure water = 22.022 kPa

Putting values in above equation, we get:

As, the mole fraction of acetone is 0.0707. This means that 0.0707 moles are present in the solution

To calculate the number of moles, we use the equation:

Molar mass of acetone = 58 g/mol

Moles of acetone = 0.0707 moles

Putting values in above equation, we get:

Hence, the mass of acetone that must be added is 4.101 grams

the correct answer is:

a. formation of precipitation